Question

Divide 56 in 4 parts in AP such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5:6

Answer 1

Solution-:

Let the four parts be (a – 3d), (a – d), (a + d), (a + 3d).

Then, sum = 56

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56

⇒ 4a = 56

⇒ a = 14

Now we will have after plugging in the 4 numbers we get
(14-d), (14-d), (14+d) and ( 14+3d)
According to condition
(14-3d)(14+3d)/(14-d)(14+d) = 5/6
(196-9d^2)/(196-d^2)=5/6
By cross multiply
6(196-9d^2)=5(196-d^2)
1176-54d^2=980-5d^2
-54d^2+5d^2=980-1176
-49d^2 =-196
d^2 = 4
d= +/- 2
So we have a= 14 and d= -2
(a-3d) =14+6=20
a-d=14+2=16
a+d=14-2=12
a+3d=14-6= 8
if we will take a= 14 and d= 2 we will get
8,12,16,20 Answer